Let $A_0, A_1, A_2$ be subsets of $\{2, 3, ..., 49\}$ whose elements give remainder $0$, $1$, $2$ after dividing by $3$. Note that $A_0 \cup A_1 \cup A_2 = \{2, 3, ..., 49\}$. Define a mapping $f : \{A_0, A_1, A_2\} \to \{0, 1, 2\}$ as follows: if $x \in A_i$ then we put $x = i$. Therefore, to form a sequence with the given condition, it is sufficient to arrange $1$'s and $2$'s so the sum of them is not divisible by $3$ and in the remaining places put $0$'s. Since $16 + 17 = 33$, consider the sequence $2, 2, 1, 2, 1, ..., 2, 1$ and we need to arrange $16$ $0$'s between them. Observe that $S_1$ is not divisible by $3$. Therefore, we don't put $0$ in the $1$st place. Number of ways putting $0$'s in the remaining $48$ places is $\frac{48!}{32! \cdot 16!}$ and number of ways distributing elements of $A_0$ in a chosen place is $16!$. Thus, by the product principle, number of ways distributing elements of $A_0$ is $\frac{48!}{32! \cdot 16!} \cdot 16! = \frac{48!}{32!}$. In the sequence $2, 2, 1, 2, 1, ..., 2, 1$ number of ways distributing $1$'s and $2$'s is $16! \cdot 17!$. Hence we have $\frac{48!}{32!} \cdot 16! \cdot 17!$.